Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
public class Main {
  	static int mod = 1000000007;
	public static void main(String args[]){
    	Scanner s = new Scanner(System.in);
      	int N = s.nextInt();
      	double[] dp = new double[N+1];
      	dp[0] = 1;
      	for(int i=1;i<=N;i++){
        	double p_head = s.nextDouble();
          	for(int j=i;j>=0;j--){
            	dp[j] = ((j==0)?0:((p_head)*dp[j-1])) + (1-p_head)*dp[j];
            }
        }
      	double ans = 0;
      	for(int i=0;i<=N;i++){
        	int t = N-i;
          	if(t<i){
            	ans+=dp[i];
              	//System.out.println(i+" "+dp[i]);
            }
        }
      	System.out.println(ans);
    }
}